Physics: Foundations of Physical Science

MGU Physics · Preparatory Volume

Chapter 36: Buoyancy

36.1 Why Objects Float or Sink

The everyday intuition about floating and sinking is roughly this: light things float, heavy things sink. This is wrong as stated. A steel aircraft carrier floats; a grain of sand sinks. The relevant quantity is not weight alone but the relationship between the weight of an object and the weight of fluid that object displaces. When an object is placed in a fluid, the fluid exerts forces on every surface of the object. These forces are pressure forces — they act perpendicular to each surface element. Because pressure increases with depth (Chapter 35), the pressure on the bottom of the object is greater than the pressure on the top. The net result of all these pressure forces is an upward force called the buoyant force. Whether the object floats or sinks depends on the competition between two forces: its weight (downward) and the buoyant force (upward). If the buoyant force equals or exceeds the weight, the object floats. If the weight exceeds the maximum possible buoyant force, the object sinks. The maximum possible buoyant force is the weight of fluid the object would displace if fully submerged.

The tendency to float or sink is determined by density, not by total mass or weight. An object whose average density is less than the fluid will float; an object denser than the fluid will sink. This is why a hollow steel ship (average density less Physics: Foundations of Physical Science Chapter 36 — Buoyancy than water) floats while a solid steel sphere (density 7870 kg/m³, much greater than water) sinks.

36.2 Pressure Differences in Fluids

Consider a rectangular block submerged in a fluid of density ρ_f. The block has height

h, horizontal cross-sectional area A, and its top face is at depth d below the surface. Its bottom face is at depth d + h.

The pressure on the top face is P_top = P₀ + ρ_f g d. The pressure on the bottom face

is P_bottom = P₀ + ρ_f g(d + h). The difference in pressure between the bottom and

top faces is:

ΔP = P_bottom −P_top = ρ_f g h

The downward force on the top face is F_top = P_top × A (pushing down). The upward

force on the bottom face is F_bottom = P_bottom × A (pushing up). The net upward

force is:

F_b = F_bottom −F_top = ΔP × A = ρ_f g h A

Note that d, the depth of the top face, cancels out entirely. The buoyant force does not depend on how deep the object is submerged — only on its volume and the fluid density. A submerged cube experiences the same buoyant force at 1 m depth as at 100 m depth (assuming the fluid density is uniform, which is an excellent approximation for liquids).

36.3 Origin of the Buoyant Force

The derivation above was for a rectangular block, but the result is completely general. For any shape, the net upward pressure force equals the weight of fluid that would occupy the volume of the object. This can be shown by imagining the object removed and replaced by an identical volume of fluid: that fluid volume would be in equilibrium, so the net pressure force on it (from the surrounding fluid) must equal its weight. The same pressure forces act on any object of the same shape and volume, regardless of what the object is made of. Physics: Foundations of Physical Science Chapter 36 — Buoyancy The buoyant force arises purely from the pressure gradient in the fluid, not from anything special about the object. It acts even on objects much denser than the fluid — the buoyant force on a steel block submerged in water is the same as the buoyant force on an identically shaped block of wood. The difference in behaviour (steel sinks, wood floats) comes from the different weights of the two blocks, not from different buoyant forces.

Archimedes of Syracuse (287–212 BC) is credited with discovering the relationship between buoyancy and displaced fluid volume. According to legend, he was asked by King Hiero II to determine whether a gold crown had been adulterated with silver without melting it down. By noticing the water rising when he entered a full bath, he realised that displaced volume provides the key. He allegedly ran through the streets shouting “Eureka” (I have found it). The method works because gold and silver have different densities: equal masses of gold and silver displace different volumes of water.

36.4 Archimedes' Principle

Archimedes' principle is the formal statement of what the pressure-difference argument shows:

Archimedes' Principle: The buoyant force on an object immersed in a fluid equals the weight of the fluid displaced by the object.

F_b = ρ_f V_displaced g

Archimedes' principle applies to all fluids — liquids and gases alike. A helium balloon displaces air; the buoyant force on it equals the weight of air displaced. The same principle that makes steel ships float in water makes hot-air balloons and dirigibles

float in air. The only difference is the magnitude of the fluid density (ρ_air

1.2 kg/m³

vs ρ_water = 1000 kg/m³), which is why air buoyancy is negligible for most solid

objects but dominant for lightweight gas-filled envelopes.

36.5 Calculating Buoyant Force

To calculate the buoyant force, identify the fluid density and the volume of fluid displaced. For a fully submerged object, the displaced volume equals the object's total volume. For a floating object, it equals only the submerged fraction of the object's Physics: Foundations of Physical Science Chapter 36 — Buoyancy volume.

Step 1: Identify the fluid: Find ρ_f (density of the fluid the object is in, not the object

itself).

Step 2: Find displaced volume:

For fully submerged: V_displaced = total volume of

object. For floating: V_displaced = submerged fraction × total volume.

Step 3: Apply formula: F_b = ρ_f V_displaced g. Units: kg/m³ × m³ × m/s² = N. ✓

has volume 0.002 m³. It is fully submerged in fresh water. F_b = ρ_f V g = 1000 ×

0.002 × 9.81 = 19.6 N. The rock’s weight = ρ_rock V g = 2700 × 0.002 × 9.81 = 53.0

N. Since weight > F_b, the rock sinks. Net downward force = 53.0

19.6 = 33.4 N.

A spherical helium balloon has

radius 0.30 m. V = (4/3)πr³ = 0.113 m³. F_b = ρ_air V g = 1.204 × 0.113 × 9.81 = 1.33

N. This is the upward force from the displaced air.

36.6 Apparent Weight in Fluids

When an object is submerged in a fluid, a scale supporting the object reads less than the object’s true weight because the fluid is partially supporting it. The reading on the scale is the apparent weight:

W_apparent = W_true −F_b = ρ_obj V g −ρ_f V g =

(ρ_obj −ρ_f) V g

Apparent weight loss equals the buoyant force. This is used in hydrostatic weighing to determine body composition: a person is weighed in air (W_air) and then fully submerged in water (W_water). The buoyant force is W_air

W_water. Since F_b =

ρ_water V g, the body volume V = (W_air

W_water)/(ρ_water g). Body density =

W_air/(V g). From body density, body fat percentage can be estimated using known densities of fat and lean tissue.

A steel wrench (density 7870 kg/m³, volume 50 cm³ =

50 × 10 ⁶ m³) is weighed while submerged in fresh water. True weight = 7870 × 50 ×

10 ⁶ × 9.81 = 3.86 N. F_b = 1000 × 50 × 10 ⁶ × 9.81 = 0.491 N. Apparent weight =

3.86

0.49 = 3.37 N. The wrench appears about 12.7% lighter in water.

Physics: Foundations of Physical Science Chapter 36 — Buoyancy

36.7 Floating Equilibrium

An object floats when the buoyant force exactly equals its weight, with the object only partially submerged. At equilibrium:

F_b = W

⇒

ρ_f V_sub g = ρ_obj

V_total g Rearranging gives the fraction of the object’s volume that is submerged:

V_sub / V_total = ρ_obj / ρ_f

Ice density = 917 kg/m³; seawater density = 1025

kg/m³. Submerged fraction = 917/1025 = 0.895 = 89.5%. About 89.5% of a floating

iceberg’s volume is below the waterline. Only 10.5% is visible above the surface — hence the expression “tip of the iceberg” for a small visible portion of a much larger hidden problem.

A pine log (density 550 kg/m³) floats in fresh water (ρ =

1000 kg/m³). Submerged fraction = 550/1000 = 0.55 = 55%. Just over half the log sits

below the waterline.

36.8 Density and Flotation Conditions

The behaviour of an object in a fluid depends entirely on the comparison of densities: Condition Behaviour

ρ_obj < ρ_f

Floats (partially submerged) Wood on water, ice in water, oil on water

ρ_obj = ρ_f

Neutral buoyancy (suspended at any depth) Cartesian diver at equilibrium, fish with adjusted swim bladder

ρ_obj > ρ_f

Sinks (buoyant force insufficient to support weight) Steel, rock, sand in water Physics: Foundations of Physical Science Chapter 36 — Buoyancy

Table 36.1

Neutral buoyancy is the condition where the average density of an object exactly matches the fluid density, so the object neither rises nor sinks. Submarines achieve neutral buoyancy by adjusting the amount of water in ballast tanks. Fish use a gas- filled swim bladder to regulate their average density. Scuba divers wear weighted belts and buoyancy control devices to achieve near-neutral buoyancy at chosen depths. The average density of a hollow object includes the air or gas inside it. A steel ship has an enormous internal air volume. Its average density (total mass divided by total enclosed volume including internal air) is less than water, so it floats. If the hull is breached and water floods in, average density increases until it exceeds seawater density, and the ship sinks.

36.9 Submerged vs. Floating Objects

For a fully submerged object, the displaced volume is fixed (equal to the object’s total volume), so the buoyant force is fixed regardless of depth. If the object’s weight exceeds this fixed buoyant force, it sinks; if the buoyant force exceeds the weight, it accelerates upward and will surface. For a floating object, the system self-adjusts: if the object is pushed down, more fluid is

displaced, increasing F_b until equilibrium is restored. If the object bobs up, less fluid is

displaced, decreasing F_b until equilibrium is restored again. This is why floating

equilibrium is stable (for a correctly shaped object) — any small perturbation generates a restoring force. A fully submerged sinking object does not experience this self-correction: as it sinks deeper, the buoyant force remains constant (because displaced volume is fixed), so the net downward force is constant and it continues sinking at a rate determined by drag.

An object will reach floating equilibrium only if its density is less than the

fluid density. If ρ_obj > ρ_f, no stable floating equilibrium exists: the object will sink to

the bottom regardless of shape (unless its shape allows it to capture air and reduce average density, as a ship’s hull does).

36.10 Stability of Floating Bodies

A floating object in equilibrium may be stable (returns to upright after a small tilt) or unstable (capsizes when tilted). Stability depends on the relationship between the centre of gravity (G) and the metacentre (M), a geometric point determined by the Physics: Foundations of Physical Science Chapter 36 — Buoyancy shape of the waterplane. The centre of buoyancy (B) is the centroid of the displaced fluid volume. When a ship tilts, B shifts to one side because the shape of the displaced volume changes. The point at which the vertical line through the new B intersects the original vertical axis of the ship is called the metacentre M. If M is above G, the ship is stable: the buoyant force creates a righting moment that restores upright. If M is below G, the ship is unstable and will capsize. This is why ship designs are tested for metacentric height (GM = distance from G to M). Cargo ships must be loaded carefully to keep the centre of gravity low enough. Overloading upper decks or uneven cargo distribution raises G toward or above M, creating capsize risk. The 1994 Estonia ferry disaster and many other maritime casualties involved loss of stability.

For high school physics, the key stability rule is: low centre of gravity improves stability. Ships, racing cars, and double-decker buses are all designed to keep heavy components as low as possible for the same reason.

36.11 Applications of Buoyancy in Engineering and Nature

Buoyancy governs an enormous range of engineering systems and natural phenomena. In naval architecture, the design of ships, submarines, and offshore platforms requires precise calculation of buoyant forces, stability, and load distribution. A submarine surfaces by blowing compressed air into ballast tanks to displace water; it submerges by flooding those tanks to increase average density above seawater. In aviation and meteorology, atmospheric buoyancy drives convection: warm air, being less dense than cool air, rises. Thunderstorms, sea breezes, and global atmospheric circulation are all buoyancy-driven. Hot-air balloons heat the air inside the envelope to reduce its density below ambient air density, generating an upward buoyant force. In medicine, hydrostatic weighing uses buoyant force to determine body volume and hence body density, from which body fat percentage is estimated (since fat is less dense than lean muscle or bone). More recently, underwater weighing has been supplemented by air displacement plethysmography (Bod Pod), which uses the same principle in air rather than water. In geology and geophysics, isostasy is the buoyancy of Earth’s crust floating on the

denser mantle (density ρ_mantle

3300 kg/m³; ρ_crust

2700 kg/m³). Mountains have deep crustal roots that extend downward into the mantle, just as icebergs extend below water. Ice ages load continents with ice, depressing the crust; after glaciers melt, the crust slowly rebounds — this post-glacial rebound is still occurring in Scandinavia Physics: Foundations of Physical Science Chapter 36 — Buoyancy at several millimetres per year. In biology, swim bladders in fish, pneumatophores in some marine invertebrates, and the gas-filled chambers of nautiluses all exploit buoyancy for depth control. The nautilus fills its chambers with a gas (mostly nitrogen) at less than atmospheric pressure and regulates buoyancy by controlling the amount of fluid inside those chambers via an osmotic pump.

36.12 Common Mistakes in Buoyancy Problems

•

Using object density instead of fluid density in the buoyant force formula. F_b =

ρ_fluid V g, not ρ_object V g. The buoyant force depends on the fluid being

displaced, not the object doing the displacing. • Using total volume instead of submerged volume for a floating object. If an

object is only 60% submerged, V_displaced = 0.60 V_total. Using V_total

overestimates F_b by a factor of 1/0.60. • Forgetting that buoyant force is constant with depth for a fully submerged incompressible object. Students sometimes think a deeper object experiences greater buoyancy because pressure is greater. The net upward force from pressure is the same at any depth because both top and bottom pressures increase equally. •

Confusing weight and mass in F_b = ρ V g. The buoyant force is already a force

(in newtons). Do not multiply again by g when comparing to weight. Weight W = mg; F_b = ρ_f V g. Compare forces directly. • Incorrectly applying the floating fraction formula to a sinking object. V_sub/V_total = ρ_obj/ρ_f applies only when the object is floating at equilibrium. If ρ_obj > ρ_f, the object sinks and there is no floating equilibrium. • Ignoring the buoyancy of air for gas-filled objects. For balloons, blimps, and

similar objects, the relevant fluid is air, not water. Use ρ_air

1.204 kg/m³.

Always identify which fluid is being displaced before writing the buoyant force formula. A balloon rising in air displaces air; a ship floating on water displaces water; a submarine hull experiences the buoyancy of seawater. Using the wrong fluid density is the most common buoyancy calculation error.

Chapter Summary

Buoyancy is the net upward pressure force that a fluid exerts on any object immersed Physics: Foundations of Physical Science Chapter 36 — Buoyancy in it. It arises because fluid pressure increases with depth, producing a greater upward force on the bottom of an object than the downward force on the top. •

Archimedes' Principle: F_b = ρ_f V_displaced g. The buoyant force equals the

weight of fluid displaced. •

For a fully submerged object, V_displaced = total volume of object. F_b is

independent of depth. • Floating equilibrium: F_b = W, giving submerged fraction = ρ_obj/ρ_f. • If ρ_obj < ρ_f: floats. If ρ_obj = ρ_f: neutral buoyancy. If ρ_obj > ρ_f: sinks. •

Apparent weight in fluid = true weight

buoyant force = (ρ_obj

ρ_f) V g. • Stability of floating bodies depends on the position of the metacentre relative to the centre of gravity. • Applications: naval architecture, submarine buoyancy control, atmospheric convection, body composition measurement, geophysical isostasy, biological buoyancy organs.

Key Terms

buoyant force (F_b): The net upward force exerted by a fluid on an immersed object,

Archimedes' principle: The buoyant force on an object equals the weight of the fluid

displaced volume: The volume of fluid pushed aside by an immersed object; equals

apparent weight: The weight of a submerged object as measured by a supporting

scale; equals true weight minus buoyant force: W_app = W

floating equilibrium: The state in which the buoyant force exactly equals the weight of

neutral buoyancy: The condition in which an object’s average density equals the fluid

centre of buoyancy (B): The centroid of the displaced fluid volume; the point through

metacentre (M): The intersection point of the line of action of the buoyant force after

Physics: Foundations of Physical Science

metacentric height (GM): Distance from centre of gravity G to metacentre M; positive

isostasy: The buoyancy of Earth’s tectonic plates floating on the denser mantle; the

Concept Review Questions

Answer in complete sentences. No calculations required unless stated.

A steel ship floats, but a solid steel ball sinks. Explain this using average density.

What makes the ship’s average density different from solid steel?

State Archimedes' principle in your own words. Does it apply to gases? Give an

example.

A block of wood floats with 70% of its volume below the waterline. What is the

density of the wood? (State the formula and result; no calculator needed for the setup.)

You submerge an object in water and weigh it with a spring scale. The reading is

less than its weight in air. Explain this difference in terms of buoyant force.

A submarine at a fixed depth can maintain neutral buoyancy by pumping water

into or out of its ballast tanks. Explain the density logic: what happens to average density when water is pumped in? When it is pumped out?

An ice cube floats in a glass of water. When the ice melts completely, does the

water level in the glass rise, fall, or stay the same? Explain using buoyancy and density.

Why does the buoyant force on a fully submerged rigid object not change with

depth? Use the pressure-difference argument to justify your answer.

A ship designed for ocean travel (seawater, ρ = 1025 kg/m³) sails into a

freshwater river. What happens to the waterline? Does the ship sink lower or rise higher? Why?

Buoyancy Problems

Physics: Foundations of Physical Science Chapter 36 — Buoyancy

A solid aluminium sphere (radius 5.0 cm, ρ_Al = 2700 kg/m³) is fully submerged

in fresh water. (a) Find the volume of the sphere. (b) Find the buoyant force. (c) Find the true weight. (d) Find the apparent weight.

A block of wood (density 650 kg/m³) has dimensions 0.20 m × 0.15 m × 0.10 m.

It is placed in fresh water. (a) Find the mass of the wood. (b) Using the floating

fraction formula, find what depth the wood sinks to. (c) Verify that F_b = W at

this depth.

A 75 kg person is weighed while fully submerged in water. The scale reads 13.5

kg-force (133 N). (a) Find the buoyant force on the person. (b) Find the person’s volume. (c) Find the person’s average density.

A helium-filled balloon (radius 0.40 m, mass of envelope + He = 0.25 kg) floats

in air. (ρ_air = 1.204 kg/m³, ρ_He = 0.164 kg/m³). (a) Find the total volume. (b)

Find the mass of helium. (c) Find the buoyant force. (d) Find the total weight (envelope + He). (e) Find the net upward force. (f) What additional mass can the balloon lift?

A 20 000 tonne ship enters a freshwater river from the sea (seawater ρ = 1025

kg/m³, fresh water ρ = 1000 kg/m³). The waterplane area (cross-sectional area at

the waterline) is 2000 m². By how many centimetres does the ship sink lower when it enters fresh water? (Hint: The ship’s weight is unchanged; it must displace more freshwater volume to balance the same weight. Find the extra volume needed, then divide by waterplane area.) Floating/Sinking Analysis For each object, determine whether it floats, sinks, or achieves neutral buoyancy in the given fluid. If it floats, find the submerged fraction. If it sinks, find the net downward force and apparent weight.

Cork (ρ = 240 kg/m³, volume 20 cm³) in fresh water.
Gold bar (ρ = 19 300 kg/m³, volume 50 cm³) in mercury (ρ_Hg = 13 600 kg/m³).
Pumice (ρ = 900 kg/m³, volume 500 cm³) in seawater (ρ = 1025 kg/m³).
A hollow plastic ball: total mass 80 g, outer radius 6.0 cm, filled with air. Does it

float in fresh water? Find the fraction submerged.

A concrete block (ρ = 2300 kg/m³, volume 0.005 m³) suspended by a rope at the

bottom of a saltwater harbour. Find the tension in the rope. Physics: Foundations of Physical Science Chapter 36 — Buoyancy

Archimedes' Principle Exercises

A crown is weighed in air (weight = 5.00 N) and then submerged in water

(apparent weight = 4.55 N). (a) Find the buoyant force. (b) Find the volume of

the crown. (c) Find the average density of the crown. (d) Is the crown pure gold

(ρ_Au = 19 300 kg/m³)? If not, what does the density suggest?

A stone submerged in oil (density 850 kg/m³) has an apparent weight of 8.20 N.

Its true weight is 12.50 N. Find the volume and density of the stone.

A beaker of water sits on a scale reading 2.400 kg. A 500 g steel block (density

7870 kg/m³) is lowered into the water on a string (not touching the beaker). (a) Find the volume of the block. (b) Find the buoyant force on the block. (c) By how much does the scale reading change? Explain the direction of the change using Newton’s third law.

Applied Fluid Statics Problems

A 500 kg wooden raft (density 600 kg/m³) must carry a steel anchor (density

7870 kg/m³, mass 80 kg). (a) Find the volume of the raft. (b) Find the total weight (raft + anchor). (c) What volume of water must the raft displace to float the system? (d) Find the depth the raft sinks to if it is 3.0 m long and 2.0 m wide.

A hot-air balloon (envelope + basket + fuel mass = 900 kg) uses heated air

inside a volume of 2200 m³. Ambient air density = 1.204 kg/m³. (a) Find the

buoyant force when fully inflated. (b) What must the density of the hot air inside be so that the balloon can just lift off? (c) What temperature does this

correspond to, if ambient air is at 20°C? (Use ρ ∝1/T for an ideal gas with T in

kelvin.)

An underwater concrete dam is 60 m wide, 4 m tall, and 3 m thick. Density of

concrete = 2400 kg/m³; density of water = 1000 kg/m³. (a) Find the weight of

the dam. (b) Find the buoyant force on the dam. (c) Find the net downward force holding the dam to the riverbed. Why does a dam not float away?

Chapter 36 Checkpoint

Physics: Foundations of Physical Science Chapter 36 — Buoyancy Before moving on to Chapter 37, confirm you can do each of the following. I can… See Section Explain why objects float or sink in terms of density comparison 36.1, 36.8 Derive buoyant force from pressure differences across a submerged object 36.2–36.3

State Archimedes' principle and write the formula F_b = ρ_f V

g 36.4 Calculate buoyant force for fully submerged and floating objects 36.5 Calculate apparent weight of a submerged object 36.6 Apply the floating equilibrium condition and find the submerged fraction 36.7 Classify objects as floating, neutral, or sinking based on density 36.8–36.9 Explain what determines stability of a floating body (metacentre) 36.10 Describe engineering and natural applications of buoyancy 36.11 Avoid common errors in buoyancy problems (wrong fluid, wrong volume) 36.12